题解 | #统计每个学校各难度的用户平均刷题数#

select 
    u.university,difficult_level,count(answer_cnt)/count(distinct q.device_id) as avg_answer_cnt
from 
    user_profile u,
    question_practice_detail q,
    question_detail qd
where
    u.device_id=q.device_id and q.question_id =qd.question_id
group by
    university,qd.difficult_level
order by 
    university;